#A45 INTEGERS 9 (2009), BALANCED SUBSET SUMS IN DENSE SETS OF INTEGERS. Gyula Károlyi 1. H 1117, Hungary

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1 #A45 INTEGERS 9 (2009, BALANCED SUBSET SUMS IN DENSE SETS OF INTEGERS Gyula Károlyi 1 Institute of Mathematics, Eötvös University, Pázmány P. sétány 1/C, Budapest, H 1117, Hungary karolyi@cs.elte.hu Received: 10/31/07, Revised: 6/10/09, Accepted: 7/7/09, Published: 11/27/09 Abstract Let 1 a 1 < a 2 < < a n 2n 2 denote integers. Assuming that n is large enough, we prove that there exist ε 1,..., ε n { 1, +1} such that ε 1 + +ε n 1 and ε 1 a 1 + +ε n a n 1. This result is sharp, and in turn it confirms a conjecture of Lev. We also prove that when n is even, every integer in a large interval centered at (a 1 + a a n /2 can be represented as the sum of n/2 elements of the sequence. 1. Introduction At the Workshop on Combinatorial Number Theory held at DIMACS, 1996, Lev proposed the following problem. Suppose that 1 a 1 < a 2 < < a n 2n 1 are integers such that their sum σ = n a i is even. Assuming that n is large enough, does there exist I {1, 2,..., n} such that i I a i = σ/2? Note that a restriction has to be imposed on n, since the sequences (1, 4, 5, 6 and (1, 2, 3, 9, 10, 11 provide counterexamples otherwise. The answer is in the affirmative: It follows from a result of Lev [3], that if n is large enough, then every integer in the interval [840n, σ 840n] can be expressed as the sum of different a i s, see [1]. In this paper we prove the following strong version of Lev s conjecture. Theorem 1 Let 1 a 1 < a 2 < < a n 2n 1 denote integers such that at least one of the numbers a i is even. If n 89, then there exist ε 1,..., ε n { 1, +1} such that ε ε n 1 and ε 1 a ε n a n 1. Note that although most likely the condition n 89 can be relaxed, it is not merely technical. The sequence (1, 2, 3, 8, 9, 10, 14, 15 demonstrates that Theorem 1 is not valid with n = 8. A more intrinsic aspect of the evenness condition is that there exists an index 1 ν n 1 such that a ν+1 a ν = 1. This is certainly the case if a n 2n 2. Corollary 2 Let 1 a 1 < a 2 <... < a n 2n 2 denote integers. If n 89, then there exist ε 1,..., ε n { 1, +1} such that ε 1 + +ε n 1 and ε 1 a 1 + +ε n a n 1. 1 Visiting the CWI in Amsterdam. Research partially supported by Hungarian Scientific Research Grants OTKA T and K67676.

2 INTEGERS: 9 ( Now the conjecture of Lev follows almost immediately from the above theorem, unless a i = 2i 1 for 1 i n. Even in that case, it is easy to check that the conclusion of Theorem 1 remains valid if n 0, 1 or 3 (mod 4. This is not the case, however, if n 2 (mod 4. Indeed, let n = 4k + 2 and suppose that ε 1,..., ε n { 1, +1} such that ε ε n 1. Consider I = {1 i n ε i = +1}, then I = 2k + 1. Therefore A = i I a i and B = i I a i are odd numbers. However, A+B = n a i = (4k +2 2 is divisible by 4, hence A B 2 (mod 4, and ε 1 a ε n a n = A B 2. Nevertheless, choosing we find that I = {1, 2, 3, 5} k {4i, 4i + 1} {1, 2,..., n} i=2 a i = 1 2 i I n a i = σ 2, confirming the conjecture of Lev in this remaining case, too. The method of the proof of Theorem 1 allows us to obtain the following generalization. Theorem 3 For every ε > 0 there is an integer n 0 = n 0 (ε with the following property. If n n 0, 1 a 1 < a 2 <... < a n 2n 2 are integers, and N is an integer such that N ( εn2, then there exist ε 1,..., ε n { 1, +1} such that ε ε n 1 and ε 1 a ε n a n N 1. Consequently, every integer in a long interval can be expressed as a balanced subset sum. Corollary 4 If n is large enough and 1 a 1 < a 2 < < a n 2n 2 are integers, then for every integer k [σ/2 n 2 /24, σ/2 + n 2 /24] there exists a set of indices I {1, 2,..., n} such that I { n/2, n/2 } and i I a i = k. Proof. We apply Theorem 3 with ε = 9/100 1/12. If k = σ/2 + x is an integer in the prescribed interval, then for the integer N = 2x there exist ε 1,..., ε n { 1, +1} such that ε ε n 1 and ε 1 a ε n a n N 1. Since N = 2x σ ε 1 a ε n a n (mod 2, it follows that ε 1 a ε n a n = N, and with I = {i ε i = +1} we have I { n/2, n/2 } and a i = 1 ( n a i + 2 i I n ε i a i = σ 2 + x = k.

3 INTEGERS: 9 ( Note that all these results can be extended to sparser sequences under the assumption that the sequence contains sufficiently many small gaps. We do not elaborate on this here. Finally we note that if balancedness is not required, then the following result, anticipated in [2], is now available, see [1]. Theorem 5 Let 1 a 1 < a 2 < < a n l 2n 6 denote integers. If n is large enough, then every integer in the interval [2l 2n + 1, σ (2l 2n + 1] can be expressed as the sum of different a i s. Neither the length of this interval can be extended, nor can the bound 2n 6 be replaced by 2n The Proof of Theorem 1 First we note that it is enough to prove Theorem 1 when n is an even number. Indeed, let n be odd, and assume that the statement has been proved for n + 1. Consider the sequence b 1 = 1 < b 2 = a < < b n+1 = a n + 1 < 2(n There exist η 1,..., η n+1 { 1, +1} such that η η n+1 1 and η 1 b η n+1 b n+1 1. Since n + 1 is even, it follows that η η n+1 = 0. Let ε i = η i+1, then ε ε n = η 1 = 1, and n n ε i a i = n+1 η i+1 a i + n+1 η i = η i b i 1. Accordingly, we assume that n = 2m with an integer m 45. To illustrate the initial idea of the proof, consider the differences e i = a 2i a 2i 1 for i = 1, 2,..., m. If we can find δ 1,..., δ m { 1, +1} such that m δ ie i < 2, then the choice ε 2i = δ i, ε 2i 1 = δ i clearly gives the desired result. This is the case, in fact, when m e i 2m 2, as it can be easily derived from the following two simple lemmas. They are intentionally formulated so that their application is not limited to integer sequences. Lemma 6 Let e 1,..., e k 1 and suppose that E = k e i βk (β 2 β

4 INTEGERS: 9 ( for some positive real number β. Then e i<s+1 e i s holds for every real number s satisfying β 1 s k β. Proof. The inequality is clearly valid if s 0. Suppose that s is a positive number satisfying e i<s+1 e i < s. Then the number of indices i such that e i < s + 1 is smaller than s. Hence the number of those i with e i s + 1 is greater than k s, therefore (s + 1(k s < E βk (β 2 β. The left-hand side is a concave function of s, attaining the value βk (β 2 β at the points β 1 and k β. Consequently, we have either s < β 1 or s > k β, proving the assertion. Lemma 7 Let e 1,..., e k 1 and suppose that e i s (1 e i<s+1 holds for every integer 1 s max{e i 1 i k}. Let F be any number such that F < k e i + 2. (2 Then there exist ε 1,..., ε k { 1, +1} such that k ε i e i F < 2, in particular F = k ε ie i if the e i s are integers and F k e i (mod 2. Proof. Without loss of generality, we may suppose that e 1 e 2 e k, so that e k < 2. The point is, that the condition allows us to construct ε 1,..., ε k sequentially so that the sequence of partial sums i j=1 ε je j oscillates about F with smaller and smaller amplitude, until it eventually approximates F with the desired accuracy. More precisely, let 0 = F, and define ε n and n recursively as follows. Let, for n = 1, 2,..., k, ε n = { 1 if n 1 0, 1 if n 1 < 0,

5 INTEGERS: 9 ( and let n = n 1 ε n e n ; then n = F ε 1 e 1 ε 2 e 2 ε n e n for every 0 n k. We prove, by induction, that n < e n e k 1 + e k + 2 (3 for n = 0, 1,..., k. This is true for n = 0. Thus, let 1 n k, and suppose that (3 is satisfied with n 1 in place of n. Assume, without loss of generality, that n 1 0. Then, by definition, e n n = n 1 + ( 1e n < e n e k + 2. Thus, to verify (3, it suffices to show that e n < e n+1 + +e k +2. This is definitely true, if e n+1 = e n or n = k. Otherwise we can write k e i = e i e i e n 1 > e n 2, e i<e n i=n+1 e i< e n proving the assertion. Letting n = k in (3, the statement of the lemma follows. The main idea of the proof of Theorem 1 is to find k m and a partition {a 1, a 2,..., a n } = k {x i, y i } {z 1,..., z n 2k } (4 such that e i = x i y i (1 i k and F = n 2k ( 1i z i satisfy the conditions of Lemma 7. Then Theorem 1 follows immediately. To achieve this we will construct the above partition so that k e i 4k 12 (resp. k e i 3k 6, (5 e i k 4 (resp. e i k 3 for i = 1, 2,..., k, (6 F k + 1, and (7 e i s for s = 1 and s = 2. (8 e i s Then an application of Lemma 6 with β = 4 (resp. with β = 3 will show that e i (1 i k and F satisfy the conditions of Lemma 7. More precisely, it follows from (5 and (8 that condition (1 holds for s k β, hence for every integer 1 s max{e i 1 i k} in view of (6. Finally, (2 follows from (7, given

6 INTEGERS: 9 ( that k e i k. Therefore, once we find a partition (4 with properties (5 (8, the proof of Theorem 1 will be complete. First we take care of the condition (8. If we take x k = a ν+1 and y k = a ν, then e k = 1. Moreover, since n 1 (a i+1 a i 2n 2, there must be an index µ {ν 1, ν, ν + 1, n}, such that a µ+1 a µ 2. Taking x k 1 = a µ+1 and y k 1 = a µ, condition (8 will be satisfied. Enumerating the remaining n 4 elements of the sequence (a i as with f i = b 2i b 2i 1 we find that m 2 1 b 1 < b 2 <... < b 2m 4 4m 1, m 2 f i = (b 2i b 2i 1 (4m 2 (m 3 = 3m + 1. (9 Since m > 21, there cannot be three different indices i with f i m 5. We distinguish between three cases. Case 1. If f i m 6 for 1 i m 2, then we can choose k = m, F = 0. Taking x i = b 2i and y i = b 2i 1 for 1 i k 2, conditions (6 and (7 are obviously satisfied, whereas (5 follows easily from (9: given that m 16. k e i m 2 f i + 3 3m + 4 4m 12, Case 2. There exist indices u, v such that m 5 f u f v. In view of (9 we have f u +f v (3m+1 (m 4 = 2m+5, and consequently m 5 f u f v m+10 and 0 f v f u 15. Therefore we may choose k = m 2, z 1 = b 2v 1, z 2 = b 2v, z 3 = b 2u, z 4 = b 2u 1. Constructing x i, y i (1 i m 4 from the remaining elements of the sequence (b i in the obvious way we find that F 15 < m 2 = k, each e i satisfies e i m 6 = k 4, and once again (9 gives k e i m 2 f i 2(m m + 14 < 4m 20 = 4k 12. Case 3. There exists exactly one index u with m 5 f u. From (9 it follows that f u (3m + 1 (m 3 = 2m + 4. We claim that there exist indices v, w different from u such that b 2w + b 2w 1 b 2v b 2v 1 f u m 2. (10

7 INTEGERS: 9 ( In that case we can choose k = m 3 and z 1 = b 2u, z 2 = b 2u 1, z 3 = b 2v, z 4 = b 2w, z 5 = b 2w 1, z 6 = b 2u 1 to have F m 2 = k + 1. Constructing x i, y i (1 i m 4 from the remaining elements of the sequence (b i in the obvious way this time we find that each e i satisfies e i m 6 = k 3, and k e i m 2 f i (m m + 7 3m 15 = 3k 6. It only remains to prove the above claim. The idea is to find v, w in such a way that f v, f w are small and at the same time b 2w b 2v lies in a prescribed interval that depends on the size of f u. It turns out that the optimum strategy for such an approach is the following. First, for any positive integer κ 2, introduce I κ = {i 1 i m 2, i u, f i κ}. Denote by x the number of indices i u for which f i > κ. Then (m 3 x + (κ + 1x m 2 f i f u (3m + 1 (m 5 = 2m + 6. Thus, κx m + 9, and m 3 x (1 1/κm 3 9/κ. We have proved the following. Claim 8 I κ κ 1 κ m 9 κ 3. In particular t = I 7 Write c 0 = 0 and let 6m i I 7 {b 2i 1, b 2i } = {c 1 < c 2 <... < c 2t 1 < c 2t }. Now we separate two subcases as follows. Case 3a. m 5 f u 2m 14. We will prove that there exist 1 i < j t such that Since we have we can argue that m 2 3 i,j = c 2j c 2i m 7. (11 1 c 2i c 2i 1, c 2j c 2j 1 7, (12 m 12 2 i,j 6 c 2j + c 2j 1 c 2i c 2i 1 2 i,j + 6 < 2m 7, and that implies (10. If there exists 1 i t 1 such that m 2 3 c 2i+2 c 2i m 7,

8 INTEGERS: 9 ( then (11 follows immediately. Otherwise we have c 2i+2 c 2i m or c 2i+2 c 2i m 6 for every integer 1 i t 1. In this way we distinguish between small gaps and large gaps in the sequence c 2, c 4,..., c 2t. The large gaps partition this sequence into blocks, where the gap between two consecutive elements within a block is always small. For such a block B = (c 2i, c 2i+2,..., c 2i, we call the length of B the quantity l(b = 2(i i. Since ( m < m 6, 2 in order to have a pair i, j with (11, it is enough to prove that at least one block has a length m/2 3. Then the smallest integer j satisfying c 2j c 2i m/2 3 will be convenient. We claim that there cannot be more than three blocks. Indeed, since every gap is at least 2, were there three or more large gaps, we would find that 4m 1 t 1 (c 2i+2 c 2i 3(m 6 + (t 32 i=0 ( 6m 30 3m , implying m 221/5 < 45, a contradiction. Since there are at most three blocks, one must contain at least t/3 different c 2i s, and thus its length ( t l(b m Given m 26, we conclude that indeed l(b m/2 3. Case 3b. 2m 13 f u 2m + 4. This time we prove that m i,j 3 2 m 21 2 holds with suitable 1 i < j t. In view of (12 this implies (13 m i,j 6 c 2j + c 2j 1 c 2i c 2i 1 2 i,j + 6 3m 15, and from that (10 follows. Similarly to the previous case, we may assume that there are only small and large gaps, which in this case means that c 2i+2 c 2i m or c 2i+2 c 2i 3 2 m 10

9 INTEGERS: 9 ( holds for every integer 1 i t 1. Given that (here we use m 44 2 ( m < m 10, it suffices to prove that there is a block B with l(b m/ If there were two or more large gaps, we would find that 4m 1 t 1 ( 3 (c 2i+2 c 2i 2 2 m 10 + (t 22 i=0 ( 6m 30 3m , implying m 221/5 < 45, a contradiction. Therefore there are at most two blocks, one of which containing at least t/2 different c 2i s. The length of that block thus satisfies ( t l(b m Since m 172/5, we find that l(b m/2 + 6, and the proof is complete. 3. The Proof of Theorem 3 Obviously we may assume that ε > 0 is small enough so that all the below arguments work. We fix such an ε and assume that n is large enough. As in the proof of Theorem 1, we may assume that n = 2m is an even number. Put c = 1/5 2ε. We will prove that there exists an integer k (1 cm 7 and a partition in the form (4 such that for e i = x i y i (1 i k and F = N + n 2k ( 1i z i the following conditions hold: k e i 4k 12, (14 e i (1 cm 11 k 4 for i = 1, 2,..., k, (15 F (1 cm 6 k + 1, and (16 e i s for s = 1 and s = 2. (17 e i s As in the proof of Theorem 1, we can apply Lemma 6 with β = 4, and then Lemma 7 gives the result. Clearly there exist 1 µ, ν n 1, µ {ν 1, ν, ν + 1} such that a ν+1 a ν = 1 and a µ+1 a µ 2. Putting x 1 = a ν+1, y 1 = a ν, x 2 = a µ+1, y 2 = a µ then takes care of (17. Enumerate the remaining n 4 elements of the sequence (a i as 1 b 1 < b 2 <... < b 2m 4 4m 2.

10 INTEGERS: 9 ( Take q = cm. Since q (b 2m 3 i b i q (2m 2i 3 = 2qm q(q + 4 > 2cm 2 (cm + 1(cm + 5 = (2c c 2 m 2 (6cm + 5 > ( ( 9 5 ε 4ε2 m 2 2m > 25 4ε m 2 N and b 2m 3 i b i 4m 3 for every i, there exists an integer 0 r < cm + 1 such that N sgn(n r (b 2m 3 i b i 2m 2, where sgn(n = +1, if N 0 and sgn(n = 1, if N < 0. Consider r + 1 b r+1 < b r+2 <... < b 2m 4 r 4m 2 r, and let f i = b r+2i b r+2i 1 for 1 i m 2 r, then m r 2 f i ( (4m 2 r (r + 1 (m r 3 3m. (18 Were there 3 or more indices i with f i > (1 cm 11, it would imply m r 2 f i > 3 ( (1 cm 11 + (m r 5 > (4 4cm 39 > 3m, a contradiction, if m is large enough. Thus there exist an integer s {0, 1, 2} and indices i 1,..., i s such that f i > (1 cm 11 if and only if i {i 1,..., i s }. Moreover, if s 1, then for each j {1,..., s} we have f ij 3m (m r 3 < (2 + cm + 4. Consequently, there exist δ 1,..., δ s { 1, +1} such that N sgn(n r (b 2m 3 i b i s δ j f ij < (2 + cm + 4. (19 j=1 Put κ = 3/ε (1 cm 11 and introduce I κ = {i 1 i m r 2, f i κ}.

11 INTEGERS: 9 ( Denoting by x the number of indices i with f i > κ we have (m r 2 x + (κ + 1x implying κx < (2 + cm + 3, and thus ( t = I κ = m r 2 x > Write c 0 = 0 and let 1 c 2 + c κ m r 2 f i 3m, m 3 3 κ > ( ε m. i I κ {b r+2i 1, b r+2i } = {c 1 < c 2 <... < c 2t 1 < c 2t }. We prove that there exist 1 i 1 < j 1 t such that 2 5 m 1 = c 2j1 c 2i1 4 m. (20 5 This follows immediately if there exists 1 i t 1 such that otherwise we have 2 5 m c 2i+2 c 2i 4 5 m, for every integer 1 i t 1. c 2i+2 c 2i < 2 5 m or c 2i+2 c 2i > 4 5 m Gaps in the sequence c 2, c 4,..., c 2t, which are larger than 4m/5, partition this sequence into blocks, where the gap between two consecutive elements within a block is always smaller than 2m/5. We claim that there cannot be more than three such blocks. Were there on the contrary at least three large gaps, we would find that t 1 4m 2 (c 2i+2 c 2i > 3 4 m + (t 3 2 > (4 + 2εm 6, 5 i=0 a contradiction. Now one of the blocks must contain at least t/3 different c 2i s, and thus its length satisfies ( t l(b > 2 5 m. Consequently, (20 holds with suitable elements c 2i1, c 2j1 of B. Removing i 1, j 1 from I κ and repeating the argument we find 1 i 2 < j 2 t such that {i 2, j 2 } {i 1, j 1 } = and 2m/5 2 = c 2j2 c 2i2 4m/5. Since for α = 1, 2 we have we can argue that 1 c 2iα c 2iα 1, c 2jα c 2jα 1 κ, (21 2 α κ + 1 Γ α = c 2jα + c 2jα 1 c 2iα c 2iα 1 2 α + κ 1,

12 INTEGERS: 9 ( that is, 4 5 m 3 ε < Γ α < 8 5 m + 3 ε. (22 In view of (19 and (22, there exist an integer p {0, 1, 2} and η 1,..., η p { 1, +1} such that N sgn(n r (b 2m 3 i b i s δ j f ij j=1 p α=1 4 η α Γ α < 5 m + 3 (1 cm 6. 2ε Consequently, we can choose k = m r s 2p > (1 cm 7, and the elements of the set r s p {b i, b 2m 3 i } {b r+2ij, b r+2ij 1} {c 2iα, c 2iα 1, c 2jα, c 2jα 1} j=1 α=1 can be enumerated as z 1,..., z n 2k so that F = N + n 2k ( 1i z i satisfies (16. Since f i (1 cm 11 holds for every 1 i m r 2, i {i 1,..., i s }, removing z 1,..., z n 2k from the sequence b 1,..., b 2m 4, the rest can be rearranged as x 3, y 3,..., x k, y k such that 1 e i = x i y i satisfies (15. Finally, it follows from (18 that k e i m r 2 f i + 3 3m + 3 (4 4cm 40 4k 12, therefore condition (14 is also fulfilled. This completes the proof of Theorem 3. Acknowledgment I am thankful to Vsevolod Lev who introduced me to the subject and whose suggestions simplified the presentation of this paper. I also gratefully acknowledge the hospitality and support of the CWI in Amsterdam, and of the Institute for Advanced Study in Princeton where the main part of this research was done with the support of the Alfred Sloan Foundation, NSF grant DMS and by DIMACS under NSF grant STC References [1] G. Károlyi, Long arithmetic progressions in subset sums and a conjecture of Alon, submitted. [2] V.F. Lev, On consecutive subset sums, Discrete Math. 187 (

13 INTEGERS: 9 ( [3] V.F. Lev, Blocks and progressions in subset sum sets, Acta Arithm. 106 (